when f(x)=cos(x)(1+sin(x))^2 find the average rate of change of [0, pi]

Question

anonymous · Answer

$$\frac{1}{\pi -0}\int\limits_0^{\pi} \cos(x)(1+\sin(x))^2 dx$$
Then make a u substitution on 1+sin(x) and integrate.

anonymous · Answer

i got cos^3(x) does that sound right?

anonymous · Answer

It shouldn't be a function it should be a value. Let:
$$\psi = 1+\sin(x) \implies d \psi = \cos(x) dx \implies \frac{1}{\pi} \int\limits_1^1 \psi^2 d \psi=0$$
So this tells us the average value is zero.

anonymous · Answer

Confirmation: http://www.wolframalpha.com/input/?i=integral+of+cos%28x%29%281%2Bsin%28x%29%29^2%2Cx%2C0%2Cpi

anonymous · Answer

when i integrated 1+sinx i got cosx and then i plugged it back into the equation. and combined the cosx and (cosx)^2

anonymous · Answer

It doesn't work that way...

anonymous · Answer

ok

anonymous · Answer

$$\int\limits f(x)g(x)dx \ne \int\limits f(x)dx \int\limits g(x)dx$$

anonymous · Answer

Also $$\int\limits 1+\sin(x) dx=x-\cos(x)+C$$

anonymous · Answer

so i you the chain rule?

anonymous · Answer

? You're integrating there is no chain rule.

anonymous · Answer

I'm not seeing it. i will come back to this problem later.